The composite scores of students on the ACT college entrance examination in a recent year had a Normal distribution with mean µ = 20.4 and standard deviation = 5.8.
1) What is the probability that a randomly chosen student scored 24 or higher on the ACT?
2) What are the mean and standard deviation of the average ACT score for an SRS of 30 students?
3) What is the probability that the average ACT score of an SRS of 30 students is 24 or higher?
4) Would your answers to 1, 2, or 3 be affected if the distribution of ACT scores in the population were distinctly non-Normal? Explain.
1) Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
2) Mean is the same, but SEm (standard error of the mean) = SD/√(n-1)
3) Use SEm fro #2 and the same table.
4) Table can only be used for normal distributions.
1) To find the probability that a randomly chosen student scored 24 or higher on the ACT, we can use the standard Normal distribution table or a calculator with a Normal distribution function.
First, we need to standardize the score using the formula:
z = (x - µ) / σ
where x is the score, µ is the mean, and σ is the standard deviation.
In this case, we want to find the probability of scoring 24 or higher, so we calculate the z-score for x = 24:
z = (24 - 20.4) / 5.8 ≈ 0.6207
Next, we look up the area under the standard Normal curve to the right of z = 0.6207. The cumulative probability is approximately 0.2689.
Therefore, the probability that a randomly chosen student scored 24 or higher on the ACT is approximately 0.2689, or 26.89%.
2) The mean of the average ACT scores for a simple random sample (SRS) of 30 students can be calculated using the same mean as the population, which is µ = 20.4.
The standard deviation of the average ACT scores for an SRS of 30 students, often referred to as the standard error of the mean (SEM), can be calculated using the formula:
SEM = σ / √n
where σ is the population standard deviation and n is the sample size.
In this case, the sample size is n = 30 and the population standard deviation is σ = 5.8. Therefore, the standard error of the mean is:
SEM = 5.8 / √30 ≈ 1.0607
So, the mean of the average ACT scores for an SRS of 30 students is 20.4, and the standard deviation (standard error of the mean) is approximately 1.0607.
3) To find the probability that the average ACT score of an SRS of 30 students is 24 or higher, we can use the standard Normal distribution once again, but this time with the mean and standard deviation of the average scores from the previous question.
First, we need to standardize the score using the formula mentioned earlier:
z = (x - µ) / SEM
For x = 24, µ = 20.4, and SEM ≈ 1.0607, we calculate:
z = (24 - 20.4) / 1.0607 ≈ 3.3964
Next, we find the cumulative probability associated with z = 3.3964. It is almost 1, indicating a very small probability.
Therefore, the probability that the average ACT score of an SRS of 30 students is 24 or higher is extremely low.
4) Yes, the answers to 1, 2, and 3 would be affected if the distribution of ACT scores in the population were distinctly non-Normal. The Normal distribution assumption is necessary to make valid probabilistic calculations based on the z-score formula and the Normal distribution table.
If the distribution is significantly non-Normal, such as heavily skewed or has extreme outliers, the z-score calculations and assumptions of the Normal distribution may not be appropriate. In such cases, alternative methods or non-parametric statistics may be required to assess probabilities and conduct analyses accurately.